Question

How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid?

Unbalanced equation: H2SO4 + NaOH → H2O + Na2SO4

Show, or explain, all of your work along with the final answer.

Answer 1

How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid?

Unbalanced equation: $\mathrm{H}_2\mathrm{SO}_4 + \mathrm{NaOH} \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{Na}_2\mathrm{SO}_4$

Show, or explain, all of your work along with the final answer.

Solution:

$$\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{Na}_2\mathrm{SO}_4$$

Find the amount of the substance sodium hydroxide:

$$n(\mathrm{NaOH}) = \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})}$$ $$n(\mathrm{NaOH}) = \frac{65.5g}{40g/\mathrm{mole}} = 1.64\ \mathrm{mole}$$

Find the equation of the reaction mass of water which formed. Since sulfuric acid is reacted in excess, we find that the mass of water by weight sodium hydroxide:

$$\frac{n(\mathrm{NaOH})}{n(\mathrm{H}_2\mathrm{O})} = \frac{2}{2} = \frac{1}{1}$$ $$\frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} = \frac{m(\mathrm{H}_2\mathrm{O})}{M(\mathrm{H}_2\mathrm{O})}$$ $$m(\mathrm{H}_2\mathrm{O}) = \frac{m(\mathrm{NaOH}) \cdot M(\mathrm{H}_2\mathrm{O})}{M(\mathrm{NaOH})}$$ $$m(\mathrm{H}_2\mathrm{O}) = \frac{65.5g \cdot 18g/\mathrm{mole}}{40g/\mathrm{mole}} = 29.475g$$

Answer: 29.475 g of water.